Electrostatic Potential At A Point Due To Electric Dipole

Two charges –q at A and +q at B separated by a small distance 2d constitute an electric dipole and its dipole moment is p (Fig.).

Fig. Potential due to a dipole

Let P be the point at a distance r from the midpoint of the dipole O and θ be the angle between PO and the axis of the dipole OB. Let r1 and r2 be the distances of the point P from +q and –q charges respectively.

Potential at P due to charge (+q) = q/4πε0 r1

Potential at P due to charge (−q) = -q/4πε0 r2

Total potential at P due to dipole is, V = q/4πε0 r1 + (-q/4πε0 r2)

V = q/4πε0 (1/r1- 1/r2)        ...(1)

Applying cosine law, 

r1^2 = r^2 + d^2 – 2rd cos θ

r1^2 = r^2 (1+ d^2/r^2 – 2d cos θ / r )

Since d is very much smaller than r, d^2/r^2 can be neglected

  ∴  r1 = r (1-2d cos θ / r)^1/2 

or   1/r1 = 1/r (1-2d cos θ / r)^(-1/2)

Using the Binomial theorem and neglecting higher powers,

 ∴ 1/r1 = 1/r (1+d cos θ / r)       ....(2)

Similarly,

     r2^2 = r^2 + d^2 – 2rd cos (180 – θ)

or  r2^2 = r^2 + d^2  + 2rd cos θ

     r2 = r (1+2d cos θ / r)^1/2     ( since, d^2/r^2 is negligible)

 or 1/r2 = r (1+2d cos θ / r)^(-1/2)

Using the Binomial theorem and neglecting higher powers,  

∴ 1/r2 = 1/r (1-d cos θ / r)       ....(3)

Substituting equation (2) and (3) in equation (1) and simplifying

V = q/4πε0 (1/r1- 1/r2)

V = q/4πε0 r (1+d cos θ / r - 1+d cos θ / r)

∴  V = 2qd cos θ / 4πε0 r^2

V = p cos θ / 4πε0 r^2         (here,  p = q2d)    ....(4)

Special cases :

1. When the point P lies on the axial line of the dipole on the side

of +q, then θ = 0,

∴  V = p / 4πε0 r^2  

2. When the point P lies on the axial line of the dipole on the side

of –q, then θ = 180

∴  V = -p / 4πε0 r^2 

3. When the point P lies on the equatorial line of the dipole, then,

θ = 90,

∴  V = 0